Show that $f$ is uni. cont. on $(a,b)$ if and only if it is continuous $(a,b)$ and $\lim\limits_{x\to a^+}f(x)$ and $\lim\limits_{x\to b^-}f(x)$ exist

Question

I want to show $f$ is uniformly continuous on $(a,b)$ if and only if it is continuous and $\lim\limits_{x\to a^+}f(x)$ and $\lim\limits_{x\to b^-}f(x)$ exist. I saw the if statement on Show f is uniformly continuous on $(a,b)$ if it is continuous and $\lim\limits_{x\to a^+}f(x)$ and $\lim\limits_{x\to b^-}f(x)$ exist but I want to show if and only if. Here is what I've done:

Assuming $f$ is continuous on $(a,b)$ and $\lim\limits_{x\to a^+}f(x)$ and $\lim\limits_{x\to b^-}f(x)$ exist, we want to show that $f$ is uniformly continuous on $(a,b)$.

Proof from the right

Let $\{x_n\}$ be an arbitrary sequence in $(a,b)$ and $x_0\in(a,b)$ be arbitrary s.t. $\{x_n\}\to x_0$ as $n \to \infty$.

$\lim\limits_{x\to a^+}f(x):=l\;\;\text{for some}\;\; l\in\Bbb{R}$ exists $\implies$ If $\{a_n\}$ is an arbitrary sequence in $(a,a+\delta)$ s.t. $a_n\to a$ as $n \to \infty$, then $f(a_n)\to l$ as $n \to \infty$.

$\lim\limits_{x\to b^-}f(x):=k\;\;\text{for some}\;\; k\in\Bbb{R}$ exists $\implies$ If $\{b_n\}$ is an arbitrary sequence in $(b-\delta,b)$ s.t. $b_n\to b$ as $n \to \infty$, then $f(b_n)\to k$ as $n \to \infty$.

By continuity of $f$, we have that $x_n\to x_0$ as $n \to \infty\implies f(x_n)\to f(x_0)$ as $n \to \infty$. Hence, $f$ is uniformly continuous on $(a,b)$

My question is: I'm I right? Different kinds of sequential proofs, even $\epsilon-\delta$ proofs and criticisms are welcome!

Answer 1

Let $f:(a,b)\to\mathbb R$ be a function such that $\lim_{x\to a+}f(x)$ does not exist.

It is our aim to prove that $f$ is not uniformly continuous on $(a,b)$.

The absence of the limit implies the existence of a strict monotonically decreasing sequence $(x_k)_k$ in $(a,b)$ with $\lim_{k\to\infty} x_k=a$ such that $\lim_{k\to\infty} f(x_k)$ does not exist.

Then $(f(x_k))_k$ is not a Cauchy-sequence, so some $\epsilon>0$ exists such that for every $n$ we can find integers $m,k$ with $m,k>n$ and $|f(x_k)-f(x_m)|>\epsilon$.

This tells us that $f$ is not uniformly continuous on $(a,x_n]$ hence is not uniformly continuous on $(a,b)$.

Answer 2

I would prove that statement as follows: define$$\begin{array}{rccc}F\colon&[a,b]&\longrightarrow&\mathbb R\\&x&\mapsto&\begin{cases}\lim_{x\to a^+}f(x)&\text{ if }x=a\\f(x)&\text{ if }x\in(a,b)\\\lim_{x\to b^-}f(x)&\text{ if }x=b.\end{cases}\end{array}$$Then $F$ is continuous and therefore uniformly continuous. Since $f=F|_{(a,b)}$, $f$ is uniformly continuous too.

Answer 3

If you need a more direct proof, you can use the following argument.

Let $(x_n)$ be a sequence converging to $a$. We claim that $(f(x_n))$ is a Cauchy sequence. Indeed, take $\varepsilon > 0$. By uniform continuity, there is $\delta > 0$ such that for any $x, \tilde{x} \in (a, b)$, if $\lvert x - \tilde{x} \rvert < \delta$ then $\lvert f(x) - f(\tilde{x}) \rvert < \varepsilon$. The sequence $(x_n)$ is a Cauchy sequence: consequently there exists $N$ such that $\lvert x_n - x_m \lvert < \delta$ for all $m, n \ge N$, which gives that $\lvert f(x_n) - f(x_m) \lvert < \varepsilon$ for all $m, n \ge N$. Consequently, from the claim it follows that the sequence $(f(x_n))$ converges to some real number.

Now, for any two sequences, $(x_n)$, $(\tilde{x}_n)$, converging to $a$, the sequence $$ \bar{x}_n = \begin{cases} x_k & \text{ if } n = 2k - 1 \\ \tilde{x}_k & \text{ if } n = 2k \end{cases} $$ converges to $a$, and $(x_n)$, $(\tilde{x}_n)$ are its subsequences. Consequently, both $(f(x_n))$ and $(f(\tilde{x}_n))$ are subsequences of $(f(\bar{x}_n))$, so all three of them must have the same limit. This completes the proof of the fact that $\lim\limits_{x \to a^+}f(x)$ exists (as a real number).